How many grams of na2so4 are in solution?

Question

A solution containing Na₂SO₄ was treated with 0.250 M BaCl₂ solution until all the sulfate ion had reacted to form BaSO₄. The net reaction

Ba²⁺(aq) + SO₄^2-(aq) → BaSO₄(s)

required 32.3 mL of the BaCl₂ solution. How many grams of Na₂SO₄ were in the solution?

Answer 1

From the net ionic equation

Ba²⁺(aq) + SO₄^2-(aq) ==> BaSO₄(s) we see that 1 mole Ba²⁺ reacts with 1 mole SO₄^2- to -> 1 mol BaSO₄

1. Find moles of Ba²⁺ used: 0.250 moles/L x 0.0323 L = 0.008075 moles Ba²⁺
2. Find moles SO₄^2- present: 0.008075 moles Ba²⁺ x 1 mol SO₄^2-/1 mol Ba²⁺ = 0.008075 mol SO₄^2-
3. Find mass of Na₂SO₄ present: 0.008075 mol SO₄^2- x 1 mol Na₂SO₄/1 mol SO₄^2- x 142.04 Na₂SO₄/mole = 1.14698 g = 1.15 g Na₂SO₄ (to 3 significant figures)